Php mysql - get the values of last days and total of entries - php

Question

I have to get the total values of last days into an array.
First, i have an array to get the last days:
$today = date('Y-m-d');
$ldays = [];
$db = 6;
for($i=-$db; $i<=0; $i++) {
array_push($ldays, date('Y-m-d', strtotime("$i days")));
}
//this will produce an array like this:
array (size=7)
0 => string '2018-12-21' (length=10)
1 => string '2018-12-22' (length=10)
2 => string '2018-12-23' (length=10)
3 => string '2018-12-24' (length=10)
4 => string '2018-12-25' (length=10)
5 => string '2018-12-26' (length=10)
6 => string '2018-12-27' (length=10)
And now, i pull the total of values of each days from the database.
SELECT COUNT(id) as total, data FROM users WHERE data >= '2018-12-21' GROUP BY data ORDER BY data ASC
The output will be an array like this:
array (size=2)
0 =>
array (size=2)
'total' => int 1
'data' => string '2018-12-21' (length=10)
1 =>
array (size=2)
'total' => int 1
'data' => string '2018-12-24' (length=10)
As you can see, the array is missing some days, so i need to cross the 2 arrays and see if a day exists in the 2 arrays:
$response = [];
for($i=0; $i<count($ldays); $i++) {
if(in_array($ldays[$i], array_column($rows_d, 'data'))) {
$response[$ldays[$i]] = 'yes';
}
else {
$response[$ldays[$i]] = 'no';
}
}
//this will make an array like so:
array (size=7)
'2018-12-21' => string 'yes' (length=3)
'2018-12-22' => string 'no' (length=2)
'2018-12-23' => string 'no' (length=2)
'2018-12-24' => string 'yes' (length=3)
'2018-12-25' => string 'no' (length=2)
'2018-12-26' => string 'no' (length=2)
'2018-12-27' => string 'no' (length=2)
But, insted of 'yes' and 'no', i want the array to be something like
array(size=7)
'2018-12-21' => int 1
'2018-12-22' => int 0
'2018-12-23' => int 0
'2018-12-24' => int 1
'2018-12-25' => int 0
'2018-12-26' => int 0
'2018-12-27' => int 0
and i keep getting an error if i do this :
if(in_array($ldays[$i], array_column($rows_d, 'data'))) {
$response[$ldays[$i]] = $rows[$i]['total'];
}
else {
$response[$ldays[$i]] = 0;
}
I guess this is because the rows array has only 2 entrys and the days array has 7. So, how can i obtain this? tks.

Answer 1

Rather than using in_array, use array_search combined with array_column to find the appropriate index into the $rows array instead:
if (($k = array_search($ldays[$i], array_column($rows_d, 'data'))) !== false) {
$response[$ldays[$i]] = $rows_d[$k]['total'];
}
else {
$response[$ldays[$i]] = 0;
}
Note that you have used both $rows and $rows_d in your code, I'm assuming it is supposed to be $rows_d in both cases.

Answer 2

It can looks prettier
$today = date('Y-m-d');
$ldays = [];
$db = 6;
for($i=-$db; $i<=0; $i++) {
$ldays[date('Y-m-d', strtotime("$i days"))] = 0;
}
$dbResult = [['total' => 1,'data' => '2018-12-21'],['total' => 2,'data' => '2018-12-24']]
foreach ($dbResult as $v) {
$ldays[$v['data']] = $v['total'];
}
var_dump($ldays);